The vertex leafage of chordal graphs

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1 The vertex lefge of horl grphs Steven Chplik, Jurj Stho b Deprtment of Physis n Computer Siene, Wilfri Lurier University, 75 University Ave. West, Wterloo, Ontrio N2L 3C5, Cn b DIMAP n Mthemtis Institute, University of Wrwik, Coventry CV4 7AL, Unite Kingom Abstrt Every horl grph G n be represente s the intersetion grph of olletion of subtrees of host tree, so-lle tree moel of G. This representtion is not neessrily unique. The lefge l(g) of horl grph G is the minimum number of leves of the host tree of tree moel of G. The lefge is known to be polynomilly omputble. In this ontribution, we introue n stuy the vertex lefge. The vertex lefge vl(g) of horl grph G is the smllest number k suh tht there exists tree moel of G in whih every subtree hs t most k leves. In prtiulr, the se vl(g) 2 oinies with the lss of pth grphs (vertex intersetion grphs of pths in trees). We prove for every fixe k 3 tht eiing whether the vertex lefge of given horl grph is t most k is NP-omplete. In prtiulr, we show tht the problem is NP-omplete on split grphs with vertex lefge of t most k+1. We further prove tht it is NP-hr to fin for given split grph G (with vertex lefge t most three) tree moel with minimum totl number leves in ll subtrees, or where mximum number of subtrees re pths. On the positive sie, for horl grphs of lefge t mostl, we show tht the vertex lefge n be lulte in time n O(l). Finlly, we prove tht every horl grph G mits tree moel tht relizes both the lefge n the vertex lefge of G. Notbly, for every pth grph G, there exists pth moel with l(g) leves in the host tree n we esribe n O(n 3 ) time lgorithm to ompute suh pth moel. Key wors: horl grph, lefge, tree moel, lique tree, pth grph 1. Introution In the following text, grph is lwys finite, simple, unirete, n loopless. A grph G = (V, E) hs vertex set V(G) n ege set E(G). We write uv for the ege(u, v) E(G). We use N G (v) to enote the neighbourhoo of v in G, n write N G [v] = N G (v) {v}. The egree of v in G is enote by eg G (v) = N G (v). Where pproprite, we rop the inex G, n write N(v), N[v], n eg(v), respetively. We use G[X] to enote the subgrph of G inue by X V(G), n write G X for the grph G[V(G)\X]. We use G v for G {v}. We sy tht X is lique of G if G[X] is omplete grph, n X is n inepenent set of G if G[X] hs no eges. A tree moel of grph G = (V, E) is pir T = (T,{T u } u V ) where T is tree, lle host tree, eh T u is subtree of T, n pir uv is in E if n only if V(T u ) V(T v ) =. In other wors, T onsists of host tree n olletion of its subtrees whose vertex intersetion grph is G. A grph is horl if it oes not ontin n inue yle of length four or more. It is well-known [1, 7, 21] tht grph is horl if n only if it hs tree moel. Any horl grph mits possibly mny ifferent tree moels. For tree T, let L(T) enote the set of its leves, i.e., verties of egree one. If T onsists of single noe, we efine L(T) =. In other wors, we onsier suh tree to hve no leves. The lefge of horl grph G, enote by l(g), is efine s the smllest integer l suh tht there exists tree moel of G whose host tree hs l leves (see [15]). It is esy to see tht l(g) = 0 if n only if G is omplete grph, n otherwise l(g) 2. Moreover the se l(g) 2 orrespons preisely to intervl grphs (intersetion grphs of intervls of the rel line) [5]. In this sense, the lefge of horl grph G mesures how lose G is to being n intervl grph. Emil resses: hplik@s.toronto.eu (Steven Chplik), j.stho@wrwik..uk (Jurj Stho) Preprint submitte to Elsevier September 8, 2012

2 In this pper, we introue n stuy similr prmeter. Definition 1. For horl grph G = (V, E), the vertex lefge of G, enote by vl(g), is the smllest integer k suh tht there exists tree moel ( T,{T u } u V ) of G where L(Tu ) k for ll u V. In other wors, the vertex lefge of G seeks tree moel of G where eh of the subtrees (orresponing to the verties of G) hs t most k leves n the vlue of k is smllest possible. (See Figure 1 for illustrtion.) k ) e l h i f g b j ekl b) hl el eil b f b g bj ) b e l f, g, h i, j, k Figure 1: ) exmple grph G with l(g) = 4 n vl(g) = 3, b) exmple lique tree T of G, ) tree moel orresponing to (efine by) T In the subsequent text, we shll sy tht tree moel of G relizes the vertex lefge of G to inite tht the tree moel stisfies the onitions of Definition 1 for smllest possible k. Similrly, we shll sy tht tree moel of G relizes the lefge of G to inite tht the number of leves in the host tree of the tree moel is smllest possible. As in the se of lefge, the vertex lefge is nturl prmeter relte to some sublsses of horl grphs previously stuie in the literture. To see this, rell the lss of vertex intersetion grphs of pths in trees, lso known s pth grphs [8] (see lso [2, 14, 16, 18]). Now, observe tht for horl grph G, we hve vl(g) = 0 if G is isjoint union of omplete grphs, n otherwise vl(g) 2. Moreover, vl(g) 2 if n only if G is pth grph. Thus, the vertex lefge of horl grph G n be seen s wy to mesure how lose G is to being pth grph. Another onnetion omes from [11] where it is observe tht in O(kn) time one n fin: n optiml olouring, mximum inepenent set, mximum lique, n n optiml lique over of n n-vertex horl grph G with vertex lefge k if representtion of G ( tree moel relizing vertex lefge) is given. In [8] it is shown tht pth grphs n be reognize in polynomil time. Currently, the best known reognition lgorithms for pth grphs run in O(nm) time [2, 18], where n = V(G) n m = E(G). In other wors, for horl grph G, testing whether vl(g) 2 n be performe in O(nm) time. Some other restritions n vritions on the stnr tree moel hve lso been stuie. One suh fmily of these vritions is pture by the [h, s, t]-grphs (introue in [13]) efine s follows: G = (V, E) is n [h, s, t]-grph if there is tree moel ( T,{T u } u V ) of G suh tht the mximum egree of T is t most h, the mximum egree of eh of {T u } u V is s, n uv is n ege of G if n only if T u n T v hve t lest t verties in ommon. For more informtion on these grphs see [3, 10]. We summrize the results of our pper in the following theorems. Theorem 2. For every k 3, it is NP-omplete to eie, for split grph G whose vertex lefge is t most k+1, if the vertex lefge of G is t most k. Theorem 3. It is NP-omplete to eie, for n integer p n split grph G whose vertex lefge is t most 3, (i) if there exists tree moel of G in whih ll but p subtrees re pths, (ii) if there exists tree moel of G where the totl number of leves in ll subtrees is t most p. Theorem 4. For every l 2, there exists n n O(l) time lgorithm tht, given n n-vertex horl grph G with l(g) l, omputes the vertex lefge of G n onstruts tree moel of G tht relizes the vertex lefge of G. Theorem 5. There exists n O(n 3 ) time lgorithm tht, given n n-vertex horl grph G = (V, E) n tree moel(t,{t u } u V ) of G, omputes tree moel(t,{t u} u V ) of G suh tht (i) L(T u) L(T u ) for ll u V, (ii) L(T ) = l(g). 2

3 Corollry 6. For every horl grph G = (V, E), there exists tree moel(t,{t u} u V ) suh tht (i) L(T u) vl(g) for ll u V. (ii) L(T ) = l(g), In other wors, suh tree moel relizes both the lefge n the vertex lefge of G. The pper is struture s follows. In 2 we isuss some tehnil etils relte to tree moels. In 3 we present proof of Theorem 2 n then isuss how to moify this proof to obtin proof of Theorem 3. In 4 we prove Theorem 4, n in 5 we present proof Theorem 5 n Corollry 6. We lose the pper in 6 with summry n isussion of possible extensions of this work. 2. Miniml Tree Moels n Clique Trees Let G = (V, E) be horl grph. We sy tht two tree moels T = (T,{T u } u V ) n T = (T,{T u} u V ) of G re isomorphi, n write T T, if there exists n isomorphism ϕ between T n T tht inues n isomorphism between T u n T u for ll u V, nmely ϕ( V(T u ) ) = V(T u). A tree moel T = (T,{T u } u V ) of G is miniml if V(T) is smllest possible mong ll tree moels of G. A lique tree of G is tree T whose noes re the mximl liques of G suh tht for ll C, C V(T), every C on the pth between C n C in T stisfies C C C. Every lique tree T of G efines tree moel T T of G, where T T = (T,{T u } u V ) n T u is efine s T [ {C V(T) u C} ] for ll u V. Given n ege XY of tree T, we enote by T/ XY the tree obtine by ontrting XY in T. Nmely, T/ XY is the tree onstrute by removing X, Y from T, ing new vertex Z, n onneting to Z the neighbours of X n the neighbours of Y in T. For tree moel T = (T,{T u } u V ) of G n ege XY of T, by ontrting XY in T n ll subtrees T u we men the tree moel with host tree T/ XY n subtrees{t u/ XY XY E(T u )} {T u XY E(T u )}. A fmily of sets{x i } i I is si to hve the Helly property if every pirwise interseting subfmily of{x i } i I hs ommon point. In other wors, every J I suh tht X i X j = for ll i, j J stisfies i J X i =. Note tht ny olletion of subtrees of tree hs the Helly property. Ft 7. Let T = (T,{T u } u V ) be tree moel of G. Then the following sttements re equivlent. (i) T is miniml tree moel of G. (ii) T T T for some lique tree T of G. (iii) For ll XY E(T), ontrting XY in T n ll subtrees T u ontining it yiels tree moel of G = G. (iv) The mpping ψ efine for X V(T) s ψ(x) = {u V X V(T u )} is bijetion between the verties of T n the mximl liques of G. Proof. (i) (iii) n (ii) (iv) re ler, while (iii) (iv) (i) follow from the Helly property of subtrees. Note tht Ft 7(iv) sttes, in other wors, tht the set of ll verties of G whose subtrees ontin X is mximl lique of G. In prtiulr, for ny tree moel, the set of suh verties is lwys lique of G, but it is not lwys neessrily mximl lique. This is only true for miniml tree moels. Moreover, it follows from Ft 7(i) (iii) tht every tree moel (T,{T u } u V ) of G n be trnsforme (by ontrting some eges of the host tree n the subtrees) into miniml tree moel (T,{T u} u V ). Notbly, s this trnsformtion involves only ontrting eges, it follows tht this oes not inrese the number of leves both in the host tree n the subtrees, nmely L(T ) L(T) n L(T u ) L(T u) for ll u V. This observtion llows us to fous exlusively on miniml tree moels. Nmely, it shows tht if there exists tree moel with minimum number of leves in the host tree (subtrees), then there lso is miniml tree moel with minimum number of leves in the host tree (subtrees). Consequently, in the reminer of the pper, ll tree moels re ssume to be miniml tree moels unless otherwise speifie. Furthermore, using Ft 7(i) (ii), we shll view miniml tree moels of G s tree moels efine by lique trees of G. We shll swith between the two viewpoints s neee. 3

4 3. Hrness of Vertex Lefge In this setion, we first prove Theorem 2 stting tht lulting the vertex lefge of split grph is NP-omplete. To this en, we esribe polynomil-time reution from NOT-ALL-EQUAL-k-SAT; this problem is well-known to be NP-omplete [6] for k 3. We then use the sme trnsformtion for k = 2 to proue polynomil-time reution from MAX-CUT, whih will prove Theorem 3. Proof of Theorem 2. The problem is lerly in NP s one n esily ompute in polynomil time the number of leves in subtrees of given tree moel. To prove NP-hrness, we show reution from NOT-ALL-EQUALk-SAT. By stnr rguments [17], we my ssume, without loss of generlity, tht the instnes to this problem ontin no repete literls n no negte vribles. Thus we n phrse the problem s follows. NOT-ALL-EQUAL-k-SAT InstneI: olletion C 1, C 2,..., C m of k-element subsets of{v 1,..., v n }; Solution toi (if exists): set S {v 1,..., v n } suh tht eh j {1,..., m} stisfies C j S = n C j \ S =. In ition, we my ssume the following property of ny instnei. ( ) There re no istint inies i, i + suh tht v i + C j whenever v i C j. Inee, if there exist i = i + with v i + C j whenever v i C j, then we replei by nother instne I + onstrute fromi by removing v i n ll luses C j tht ontin v i. If there is solution toi, then S\{v i } is solution toi +. Conversely, if S is solution toi +, then either S is solution toi if v i + S, or S {v i } is solution toi if v i + S. Now, for the reution, we onsier n instnei stisfying ( ) n onstrut grph, enote by G I, s follows: (i) the vertex set of G I onsists of n+m+2 verties: V(G I ) = {v 1,..., v n, y 1,..., y m, z 1, z 2 }, (ii) the verties{y 1,..., y m } form lique, (iii) the verties{v 1,..., v n, z 1, z 2 } form n inepenent set, (iv) eh vertex v i is jent to ll verties y j suh tht v i C j, (v) the verties z 1, z 2 re jent to eh vertex of the lique{y 1,..., y m }. We observe tht G I is split grph with prtition into lique {y 1,..., y m } n inepenent set {v 1,..., v n, z 1, z 2 }. (See Figure 2 for n exmple of this onstrution.) C 1 = {v 1, v 2, v 3 } C 2 = {v 1, v 4, v 6 } C 3 = {v 3, v 5, v 6 } C 4 = {v 2, v 4, v 5 } Q 4 v 4 y 2 y 4 Q 2 v 2 y 1 y 4 Q 6 v 6 y 2 y 3 v 2 v 1 y 1 y 2 v 6 z 2 y 1 y 2 y 3 y 4 B z 1 z 2 z 1 y 1 y 2 y 3 y 4 A y 4 y 3 v 3 v 5 v 4 v 3 y 1 y 3 Q 3 v 5 y 3 y 4 Q 5 v 1 y 1 y 2 Q 1 y 1 y 2 y 3 y 4 z 1, z 2, v 1, v 2, v 3, v 4, v 5, v 6 Figure 2: ) the grph G I for exmple instne I with n = 6 n m = 4, b) lique tree of G I, ) orresponing tree moel of G I. We prove tht the vertex lefge of G I is: () t most k+1, n (b) is t most k if n only if there is solution toi. To o this we nlyze the liques of G I. This is esy, sine G I is split grph; ll its mximl liques re forme by tking vertex of the inepenent set with its neighbourhoo. In prtiulr, the mximl liques of G I re A = {z 1, y 1,..., y m }, B = {z 2, y 1,..., y m }, n Q i = {v i } {y j v i C j } for eh i {1,..., n}. 4

5 We first prove (). Rell tht {A, B, Q 1,..., Q n } is the set of ll mximl liques of G I, n hene, the vertex set of every lique tree of G I. Eh of the verties z 1, z 2, n v i, for i {1,..., n}, belongs to extly one of these liques, nmely A, B, n Q i, respetively. Also, eh y j, for j {1,..., m}, belongs to extly k+2 liques, nmely A, B, n {Q i1,..., Q ik } where C j = {v i1,..., v ik }. So, sine k 1, every tree spnning these liques hs t most k+1 leves. We thus onlue tht in every lique tree of G I, eh subtree orresponing to vertex of G I hs t most k+1 leves. In other wors, ny lique tree of G I ertifies tht vl(g I ) k+ 1 whih proves (). We now prove (b). Let S be solution to I. Construt tree T with vertex set {A, B, Q 1,..., Q n } n ege set {AB} {AQ i v i S} {BQ i v i S}. Let us verify tht T is lique tree of G I. Its vertex set is the set of ll mximl liques of G I. For istint i, i + {1,..., n}, the pth between Q i n Q i + ontins A or B or both, n no other vertex. Note tht Q i Q i + {y 1,..., y m } = A B. This verifies the pth between Q i n Q i +. Similrly, the pth between Q i n A or B itionlly ontins only A or B n we hve Q i A = Q i B whih verifies this pth. Tht exhusts ll pths in T n thus onfirms tht T is inee lique tree of G I. Let T T = ( T,{T v } v V(GI )) be the tree moel orresponing to T. We nlyze its subtrees. First, we onsier the subtree T vi where i {1,..., n}. As in (), we observe tht the vertex v i only belongs to one lique of G I, nmely Q i. Thus V(T vi ) = 1 implying L(T vi ) = 0 by our onvention. Similrly, the verties z 1 n z 2 eh belong to only one lique, A n B respetively, n we hve L(T z1 ) = L(T z2 ) = 0. It remins to onsier T yj for j {1,..., m}. The vertex y j belongs to the liques A, B, n k istint liques Q i1,..., Q ik where C j = {v i1,..., v ik }. The liques Q i1,..., Q ik re leves of T yj s they re leves of T. However, neither A nor B is lef of T yj. Inee, sine S is solution to I, there re inies p, r {1,..., k} suh tht v ip S n v ir S. Hene, by onstrution, T ontins eges AQ ip n BQ ir. So, T yj ontins these eges, s well s the ege AB. Thus both A n B hve t lest two neighbours in T yj n re therefore not leves of T yj. Consequently, L(T yj ) = {Q i1,..., Q ik } = k whih implies vl(g I ) k s ertifie by the tree moel T T. Conversely, suppose tht vl(g I ) k. Then there exists lique tree T of G I suh tht the orresponing moel T T = ( T,{T v } v V(GI )) stisfies L(Tv ) k for ll v V(G I ). We nlyze the struture of T. First, we observe tht AB must be n ege of T. If otherwise, the pth between A n B in T ontins some lique Q i, i {1,..., n}. As T is lique tree, we onlue {y 1,..., y m } = A B Q i = {v i } {y j v i C j }. But then v i belongs to eh C j, j {1,..., m}, n sine n k 2, there exists i + {1,..., n} ifferent from i, whih ontrits ( ). Similrly, we show tht eh Q i, i {1,..., n} is lef of T. If otherwise, some Q i hs t lest two neighbours in T. These nnot be A, B s this woul imply tringle in T, sine AB is n ege of T. Thus Q i is jent to Q i + for some i + {1,..., n}. As T is tree, we hve tht either Q i + lies on the pth from A to Q i, or Q i lies on the pth from A to Q i +. By symmetry, we my ssume the former. Thus, sine T is lique tree, we onlue {y j v i C j } = A Q i Q i + = {v i +} {y j v i + C j }. So v i + C j whenever v i C j, ontriting ( ). Now, we re rey to onstrut set S {v 1,..., v n } s follows: for eh i {1,..., n}, we put v i in S if AQ i is n ege of T. We show tht S is solution to I. If not, there exists j {1,..., m} suh tht either S C j or S C j =. We look t the subtree T yj orresponing to the vertex y j. Rell tht y j belongs to liques A, B, n k liques Q i1,..., Q ik where C k = {v i1,..., v ik }. The liques Q i1,..., Q ik re leves of T yj beuse they re leves of T (s prove bove). If S C j, we hve, by onstrution, tht A is the unique neighbour of eh of the liques Q i1,..., Q ik in T. Consequently, none of the liques Q i1,..., Q ik is jent to B in T. This shows tht B is only jent to A in T yj, n hene, is lef in T yj. But then L(T yj ) = {Q i1,..., Q ik, B} = k+1, ontriting our ssumption bout T. Similrly, if S C j =, the liques Q i1,..., Q ik re only jent to B n not to A, in whih se, A is lef of T yj leing to the sme ontrition. Therefore, S must inee be solution to I n tht onlues the proof. Proof of Theorem 3. The proof follows the sme steps s tht of Theorem 2. To keep things simple, we only esribe here the key ifferenes. Inste of NOT-ALL-EQUAL-k-SAT, we onsier the optimiztion version of the problem for k = 2, whih is known s MAX-CUT; this problem is lso known to be NP-hr (see [6]). In MAX-CUT, we re given grph H, n we seek subset S of verties of H suh tht the number of eges between S n the rest of H is mximize. To be ble to reuse our proof of Theorem 2, we st this problem s follows. MAX-CUT InstneI: olletion C 1, C 2,..., C m of 2-element subsets of{v 1,..., v n }, n n integer p Solution toi: set S {v 1,..., v n } suh tht C j S = n C j \ S = for t lest m p inies j {1,..., m} 5

6 As in the proof of Theorem 2, we my ssume tht every given instne of this problem stisfies the property ( ). (Note tht this orrespons to removing verties of egree zero n one in the orresponing input grph, whih oes not hnge hrness of the problem.) For the reution, onsier n instnei of the bove problem, nmely, olletion C 1, C 2,..., C m of 2-element subsets of {v 1,..., v n } n n integer p. Let G I enote the grph onstrute in the proof of Theorem 2 for the olletion {C 1,..., C m }. To be more speifi, G I is the grph whose vertex set is {v 1,..., v n, y 1,..., y m, z 1, z 2 } where{v 1,..., v n, z 1, z 2 } forms n inepenent set, {y 1,..., y m } forms lique, n both z 1 n z 2 re jent to ll of y 1,..., y m, while v i is jent to y j if n only if v i C j. Note tht G I is split grph. For this grph G I, repeting the rguments presente in the proof of Theorem 2, we onlue tht () vl(g I ) 3, n (b) there exists solution to I if n only if there exists tree moel of G I in whih ll but p subtrees re pths. Moreover, we my ssume tht the tree moel mentione in (b) is miniml tree moel, n thus () implies tht the totl number of leves in this moel is t most p. This follows from the ft tht eh of the verties v 1,..., v n, z 1, z 2 belongs to extly one mximl lique of G I, n hene, the subtrees orresponing to these verties hve no leves, sine the moel is miniml. This proves both prts of Theorem 3 n ompletes the proof. 4. Vertex Lefge in Boune Lefge Grphs In this setion, we isuss lulting vertex lefge in horl grphs of boune lefge. Nmely, we prove Theorem 4, tht is, for fixe l, we emonstrte how to lulte the vertex lefge of n n-vertex horl grph G with l(g) l in polynomil time, nmely, in time n O(l). We o this by enumerting lique trees of G with respet to high ( 3) egree noes. The enumertion is bse on the observtion tht the number of high-egree noes in tree is iretly relte to the number of leves. This goes s follows. For tree T, let H (T) enote the set of noes of T of egree t lest 3, n let E(T) enote the set of eges of T inient to the noes in H (T). Further, let n i enote the number of noes of egree i in T. Then ( ) H (T) = n i (i 2)n i = n 1 + (i 2)n i = n E(T) 2 V(T) = L(T) 2 i 3 i 3 E(T) (i n i ) = 2 i 3 i 3 i 1 n i + (i 2)n i = 2 H (T) + L(T) 2 3 L(T) 6 i 3 For the seon-to-lst equlity in the first line, note tht V(T) = i 1 n i while E(T) = 2 1 i 1(i n i ). We remrk tht ( ), in prtiulr, implies tht if L(T) is boune, then so re H (T) n E(T). We shll use this ft lter. Moreover we shll use the following property. Lemm 8. If T n T re two lique trees of G with E(T) = E(T ), then L(T u ) = L(T u) for ll u V(G) where T u = T [ {C V(T) u C} ] n T u = T [ {C V(T ) u C} ]. Proof. Consier vertex u V(G). Sine E(T) = E(T ), we onlue H (T) = H (T ) n eh C H (T) = H (T ) hs the sme neighbourhoo in both T n T, i.e., N T (C) = N T (C). Moreover, if noe hs egree t lest 3 in T u, then it lso hs egree t lest 3 in T, sine T u is n subgrph of T. In other wors, we hve H (T u ) H (T). In ition, we observe tht eh C V(T u ) stisfies N Tu (C) = N T (C) V(T u ), sine T u is n inue subgrph of T. By the sme token, N T u (C) = N T (C) V(Tu) for eh C V(Tu). Finlly, note tht V(T u ) = V(Tu), sine V(T) = V(T ). Thus, for eh C H (T u ), we n write N Tu (C) = N T (C) V(T u ) = N T (C) V(T u) = N T u (C). This implies C H (Tu) n eg Tu (C) = eg T u (C) for ll C H (T u ). Thus, we lulte by( ). ( L(T u ) = 2+ egtu (C) 2 ) ( 2+ egt (C) 2) = L(T C H (T u ) C H (Tu) u u) To see tht the inequlity hols, lso note tht eg T u (C) 3 for eh C H (T u), by efinition. This proves tht L(T u ) L(T u), n symmetri rgument yiels L(T u ) L(T u) whih ompletes the proof. 6

7 Now, rell tht the vertex set of every lique tree of G is the set of ll mximl liques of G. Notbly, ll lique trees hve the sme vertex set. Let C(G) enote the lique grph of G, i.e., the grph whose noes re the mximl liques of G n where two noes re jent if n only if the orresponing mximl liques interset. It is well-known [9, 19] tht every lique tree of G is spnning tree of C(G). Our lgorithm is bse on the following lemm. Lemm 9. There is n O(n 3 ) time lgorithm tht, given n n-vertex horl grph G n set F E(C(G)), eies if there exists lique tree T of G with E(T) = F n onstruts suh tree if one exists. Proof. We esribe n lgorithm for the problem s follows. Algorithm 1: Input: A horl grph G n set F E(C(G)). Output: A lique tree T of G with E(T) = F, or report tht no suh tree exists. 1 Construt grph G s follows: V(G ) = V(G) {v e e F} E(G ) = E(G) {uv e e F, e = CC, u C C } {v e v e e, e F, e e = } 2 if G is horl then 3 Construt lique tree T of G with minimum number of leves. 4 Construt tree T from T by renming eh noe C V(T ) to C V(G) 5 if T is lique tree of G n E(T) = F then 6 return T 7 return no suh tree exists We now prove orretness of the bove lgorithm. For simpliity, we shll refer to ny lique tree T with E(T) = F s solution. First, observe tht if the lgorithm returns the tree T in Line 6, then this is inee solution. This proves tht if there is no solution, the lgorithm provies the orret nswer (in Line 7). Thus, for the rest of the proof, we my ssume tht solution exists. Nmely we shll ssume there is lique tree T of G stisfying E(T ) = F. For every mximl lique C of G, efine ϕ(c) = C {v e e F, C e}. In the following lim, we isuss the properties of the grph G onstrute in Line 1. (1) G is horl, stisfies l(g ) L(T ), n ϕ is bijetion between the mximl liques of G n G. To prove the lim, we onstrut miniml tree moel of G s follows. Let T T = (T,{Tu} u V(G) ) be the miniml tree moel of G tht is efine by the lique tree T, nmely Tu = T[{C V(T ) u C}]. For eh ege e = CC F, efine Tv e = T [{C, C }]. Finlly, let T + = (T,{Tu} u V(G) {Tv e } e F }). It is esy to verify tht T + is tree moel of G. In prtiulr, eh subtree in the olletion is onnete subgrph of T. This follows from the ft tht T is lique tree of G n tht F = E(T ) E(T ). Further, for eh ege e = CC in F, we see tht the subtree Tv e intersets only subtrees Tu where C or C is in V(Tu), i.e., those where u C C. Moreover, Tv e only intersets subtrees Tv where C or C is in V(T e v ), i.e., those where e e e =. This orrespons preisely to the efinition of G. Thus, we onlue tht G is inee horl grph, n l(g ) L(T ) s T + is prtiulr tree moel of G n T is its host tree. Morever, we see tht T + is tully miniml tree moel of G. Inee, if there were tree moel of G with less thn V(T ) noes in its host tree, then by removing subtrees orresponing to the verties{v e e F} we woul obtin tree moel of G whose host tree hs less thn V(T ) noes. But this woul ontrit the minimlity of T T. This implies, by Ft 7(ii), tht there exists lique tree T + of G tht efines T +, i.e., T + = T T +. Nmely, there is n isomorphism between T + n the host tree T of T + where eh noe C V(T ) orrespons to the set of ll verties of G whose subtrees ontin C, i.e., the set {u V(G) C V(T u )} {v e e F, C V(T ve )} whih is extly ϕ(c). In other wors, V(T + ) = {ϕ(c) C V(T )}, n onsequently, ϕ onstitutes n isomorphism between T n T +. As one is lique tree of G n the other lique tree of G, we onlue tht ϕ is bijetion between the mximl liques of G n G. This proves (1). 7

8 The lim (1) shows tht the test in Line 2 suees. Now, onsier the trees T n T onstrute in Line 3 n Line 4, respetively. Note tht T is lique tree of G with L(T ) = l(g ). (2) T is lique tree of G. Rell tht T is obtine from T by renming eh noe C of T to C V(G). Moreover, by (1), the mpping ϕ is bijetion between the mximl liques of G n G. Nmely, for eh C V(T ), the set C = ϕ 1 (C ) is mximl lique of G. Therefore, we n write C V(G) = ϕ(c) V(G) = ( C {v e e F, C e} ) V(G) = C = ϕ 1 (C ). This proves tht the vertex set of T is preisely the set of mximl liques of G, n ϕ is n isomorphism between T n T, by the onstrution of T. To see tht T is inee lique tree of G, it remins to prove the onnetivity onition for T. Nmely, onsier noes C 1, C 2 V(T) n noe C 3 on the pth in T between C 1 n C 2. Sine ϕ is n isomorphism between T n T, we hve ϕ(c i ) V(T ) for i = 1, 2, 3 n ϕ(c 3 ) lies on the pth in T between ϕ(c 1 ) n ϕ(c 2 ). Thus, we onlue ϕ(c 3 ) ϕ(c 1 ) ϕ(c 2 ) beuse T is lique tree. So we write C 3 = ϕ(c 3 ) V(G) ϕ(c 1 ) ϕ(c 2 ) V(G) = C 1 C 2. This proves (2). We hve prove in (2) tht T is lique tree of G. Notbly, s T is lso lique tree of G, we onlue tht both T n T hve the sme vertex set, i.e., V(T) = V(T ). We now look t the eges of T. (3) F E(T) Consier n ege e = CC F, n rell the efinition of ϕ n the lim (1). From this it follows tht ϕ(c) n ϕ(c ) re the only mximl liques of G tht ontin v e. As ϕ(c) n ϕ(c ) re lso noes of T whih is lique tree of G, we onlue tht every mximl lique on the pth in T between ϕ(c) n ϕ(c ) lso ontins v e. But, s mentione bove, the vertex v e is in no other mximl lique of G. So this is only possible if ϕ(c) n ϕ(c ) re jent in T. Consequently, C n C re jent in T, nmely e E(T). This proves (3). (4) H (T ) H (T) n eh C H (T ) stisfies N T (C) N T (C). Consier C H (T ), nmely C is noe of T with t lest three neighbours in T. Then, by the efinition of E(T ), ll eges inient to C in T belong to E(T ). As E(T ) = F n F E(T) by (3), the eges inient to C in T re lso eges of T. In other wors, every neighbour of C in T is neighbour of C in T, nmely N T (C) N T (C). Thus C hs t lest three neighbours in T implying C H (T). This proves (4). (5) H (T) = H (T ) n E(T) = E(T ). By (4), we onlue H (T) H (T ). Now, we lulte using (1), (4), n( ) s follows. l(g ) L(T ) = 2+ C H (T ) ( egt (C) 2 ) 2+ C H (T) ( egt (C) 2) = L(T) = l(g ) Note tht the seon inequlity follows from (4) n the ft tht eg T (C) 3 for ll C H (T), while the lst equlity is byl(g ) = L(T ) n the ft tht T n T re isomorphi. Thus the inequlities in the bove formul re, in ft, equlities. Therefore, using (4), we onlue tht H (T) = H (T ) n every C H (T ) stisfies N T (C) = N T (C). To see this, rell tht eh C H (T ) ontributes to the sum on the right t lest s muh s to the sum on the left, sine N T (C) N T (C) by (4). Further, every C H (T) hs positive ontribution to the sum on the right s eg T (C) 3 by the efinition of H (T). Thus, sine the two sums re equl, the only possibility is tht H (T) = H (T ) n tht eh C H (T ) stisfies N T (C) = N T (C) s lime. To onlue the proof, rell tht E(T), resp. E(T ), is the set of eges of T, resp. T, inient to the noes in H (T), resp. H (T ). As H (T) = H (T ) n eh C H (T) = H (T ) is inient to the sme set of eges in T n T for it stisfies N T (C) = N T (C), we onlue tht E(T) = E(T ). This proves (5). Now (5) n (2) prove tht T is inee solution, nmely tht T is lique tree of G with E(T) = E(T ) = F. Hene, the test in Line 5 sues n the lgorithm orretly returns solution in Line 6. This onlues the proof of orretness of the lgorithm. To ress the omplexity, let n = V(G) s usul. First, we note tht we my ssume tht F ontins t most n 1 eges s no lique tree of G hs more thn n noes. If this is not so, we n sfely report tht no solution exists. Thus, s G hs V(G) + F = O(n) verties, we 8

9 onlue tht step 3 tkes O(n 3 ) time using the lgorithm of [12]. All other steps lerly tke t most O(n 2 ) time. Notbly, in step 2 we use liner time lgorithm from [20]. Thus the totl omplexity is O(n 3 ) s promise. Tht onlues the proof. Finlly, we re rey to prove Theorem 4. Proof of Theorem 4. Let G be horl grph with l(g) l. By Corollry 6 (proven in 5), there exists tree moel of G tht simultneously relizes both the lefge n the vertex lefge of G. By the remrks in 2, there is lso lique tree of G with this property; let T enote this lique tree. In other wors, the tree T stisfies L(T ) = l(g) n L(Tu) vl(g) for ll u V(G) where Tu = T [ {C V(T ) u C} ]. By Lemms 8 n 9, it suffies to know the set E(T ) to be ble, in polynomil time, to fin tree moel of G (possibly ifferent from T ) tht relizes the vertex lefge of G. This forms the bsis of our lgorithm s follows. Let F E(C(G)). If there exists lique tree T with E(T) = F, efine α F = mx u V(G) L(T u ) where T u = T [ {C V(T) u C} ]. If suh tree oes not exist, efine α F = +. Note tht the vlue of α F is wellefine, sine by Lemm 8 it is inepenent of the prtiulr hoie of the lique tree T. Thus, the vlue of α F n be etermine, for ny given F, in time O(n 3 ) using Lemm 9. In prtiulr, α E(T ) = vl(g) by the hoie of T. Our lgorithm tries ll possible sets F C(G) of size t most 3l 6 s nites for E(T ) n hooses one tht tht minimizes α F. If F opt is this set, the lgorithm outputs lique tree T opt of G with E(T opt ) = F opt. We lim tht this proeure orretly fins lique tree of G tht relizes the vertex lefge of G. By ( ), we observe tht E(T ) 3 L(T ) 6 3l 6. Thus, the lgorithm must, t some point, onsier E(T ) s the set F. For this set, we hve α F = α E(T ) = vl(g). By the minimlity of F opt, we onlue α Fopt α E(T ) = vl(g). Hene, α Fopt < n so the tree T opt exists. Moreover, α F vl(g) for ll sets F, by the efinition of vl(g) n α F. Thus, we must onlue α Fopt = vl(g) n onsequently by Lemm 8, the tree T opt is lique tree of G tht relizes the vertex lefge of G. This proves the orretness of our lgorithm. Finlly, let us nlyze the omplexity. Let n = V(G) s usul. Rell tht G hs t most n mximl liques. Thus there re t most n 2 eges in C(G), n hene, t most n 6l 12 hoies for the set F. For eh hoie of F, we use Lemm 9 to fin lique tree T with E(T) = F if it exists. This tkes O(n 3 ) for eh F, inluing the lultion of α F. Altogether, the running time is O(n 6l 9 ) = n O(l) s promise. Tht onlues the proof. 5. Vertex Lefge with Optimum Lefge In this setion, we prove Theorem 5 n Corollry 6. Nmely, we emonstrte tht the lgorithm from [12], solving the lefge problem, stisfies the lim of Theorem 5. This lgorithm, given horl grph G, outputs lique tree of G with minimum possible number of leves. This is one by strting from n rbitrry lique tree T of G, n itertively eresing the number of leves of T s long s possible. We observe (n formlly prove lter in this setion) tht this proess hs the itionl property tht it never inreses the number of leves in the subtrees of the tree moel T T efine by T. In other wors, if T is the lique tree resulting from this proess, then T = T T stisfies the lim of Theorem 5. This will imply tht if the strting lique tree T relizes the vertex lefge of G, then T = T T stisfies the lim of Corollry 6. For the proof of the bove, we nee to explin the inner workings of the lgorithm from [12]. This lgorithm, in ple of lique trees, opertes on the so-lle token ssignments efine s follows. For horl grph G, token ssignment of G is funtion τ tht ssigns to every mximl lique C of G, multiset τ(c) of subsets of C. We use the wor token for the members of τ(c). Note tht the sme subset my pper in τ(c) mny times. We fous on speil token ssignment tht rise from lique trees. The token ssignment efine by lique tree T of G, n enote by ε T, ssigns to every mximl lique C of G, the multiset ε T (C) = {C C CC E(T)}. In other wors, ε T (C) onsists of the intersetions of C with its neighbours in T. A token ssignment τ is relizble if there is lique tree T of G suh tht τ = ε T. (See Figure 3 for n illustrtion of these onepts.) Notie tht the token ssignment τ = ε T ontins ll the informtion neee to etermine the number of leves in T n lso the number of leves in the subtrees of the orresponing moel T T. We summrize this s follows. 9

10 e ) f k g j h b i e b) f k g b j h bi ) b b Figure 3: ) Exmple horl grph G, b) lique tree T of G, ) token ssignment τ = ε T. Lemm 10. Let G be horl grph, let T be lique tree of G, n let T T = ( T,{T u } u V(G) ) enote the tree moel of G efine by T. Let τ = ε T, n efine τ u (C) = {S S τ(c), u S} for eh u V(G). Then eg T (C) = τ(c) for ll C V(T), n eg Tu (C) = τ u (C) for ll u V(G) n ll C V(T u ). { } { } Consequently, L(T) = C τ(c) = 1 n L(T u ) = C τ u (C) = 1 for ll u V(G). In prtiulr, while there n be multiple lique trees efining the sme token ssignment, these lique trees will hve the sme sets of leves n onsequently we o not nee to istinguish them from one nother. In other wors, it suffies to mintin tht the token ssignment we onsier orrespons to some lique tree of G. This n be teste esily by pplying four prtiulr onitions s esribe in [12]. As we o not use this test here iretly, we omit further etils. (For more, see [12, Theorem 6].) Now, we re finlly rey to explin the min steps of the lgorithm from [12]. The lgorithm is given horl grph G n lique tree T of G. It strts by onstruting the token ssignment τ = ε T. Then it proees itertively. During eh itertion step, urrent token ssignment τ is exmine to etermine if there exists ifferent token ssignment orresponing to lique tree with fewer leves. This is one by heking for n ugmenting pth in τ, whih is speifi sequene of token moves (see efinitions below). If n ugmenting pth exists, we pik the shortest suh pth n exhnge tokens long the pth. This results in new token ssignment τ tht orrespons to lique tree with fewer leves. If no ugmenting pth exists, we rrive t n optiml solution (i.e., token ssignment whose orresponing lique trees ll hve l(g) leves) n we output this solution. We summrize the bove proeure s Algorithm 2. Below we provie the missing efinitions. Let G be horl grph n τ be token ssignment of G. A token move is n orere triple (C 1, C 2, S) where C 1, C 2 re mximl liques of G n S τ(c 1 ). For token move(c 1, C 2, S), we write τ (C 1, C 2, S) to enote the token ssignment τ tht is the result of moving S from τ(c 1 ) to τ(c 2 ). Nmely 1, we hve τ (C 1 ) = τ(c 1 )\{S} n τ (C 2 ) = τ(c 2 ) {S}, while τ (C) = τ(c) for ll other C {C 1, C 2 }. A sequene of token moves(c 1, C 2, S 1 ), (C 2, C 3, S 2 ),..., (C k 1, C k, S k 1 ) where k > 1 is n ugmenting pth of τ if τ(c k ) = 1 n eh j {1,..., k 1} stisfies { (i) τ (C j, C j+1, S j ) is relizble token ssignment 2 3 if j = 1, n (ii) τ(c j ) = 2 otherwise See Figure 4 for n exmple of n ugmenting pth of token ssignment τ n its pplition to τ. It is esy to see tht the pplition of n ugmenting pth ereses the number of leves in the resulting token ssignment. This, however, oes not gurntee tht the resulting ssignment orrespons to lique tree of G. Fortuntely, it n be prove tht shortest ugmenting pth hs this property, n moreover, there lwys exists n ugmenting pth unless τ orrespons to n optiml lique tree. The etils n be foun in [12]. We only remrk the following invrint whih is mintne throughout the lgorithm. 1 Note tht s both τ(c 1 ) n τ (C 1 ) re multisets, to obtin τ (C 1 ) we only remove one instne of S from τ(c 1 ) in se S ppers in τ(c 1 ) severl times. This is onsistent with the semntis of the set ifferene for multisets. 2 i.e., it orrespons to lique tree of G. 10

11 ) b b b) b b ) b b Figure 4: ) token ssignment τ (see Figure 3), b) ugmenting pth (b, f, ), ( f, k, ) irete eges, ) τ fter pplying the pth. Algorithm 2: Lefge(G,T) Input: A horl grph G, n lique tree T of G. Output: A lique tree T of G with L(T ) = l(g). 1 Initilize τ ε T /* initilize the token ssignment with the given lique tree. */ 2 while there exists n ugmenting pth of τ o 3 Let(C 1, C 2, S 1 ),...,(C k 1, C k, S k 1 ) be shortest ugmenting pth of τ 4 for ll i from 1 to k 1 o 5 τ τ (C i, C i+1, S i ) 6 return T where ε T = τ. Lemm 11. [12] In Line 2 of Algorithm 2, the vrible τ is relizble token ssignment. After this introution, we re rey to prove Theorem 5. Proof of Theorem 5. We prove the theorem by showing tht eh pplition of n ugmenting pth in Algorithm 2 oes not inrese the number of leves in the subtrees of the orresponing tree moel. In other wors, let τ be the token ssignment onsiere t the strt of some itertion (Lines 2-5) of Algorithm 2, n let (C 1, C 2, S 1 ),..., (C k 1, C k, S k 1 ) be the shortest ugmenting pth of τ onsiere in this itertion (Line 3). Let τ enote the vlue of τ fter pplying the token moves of this pth (Lines 4-5). By Lemm 11, both τ n τ re relizble token ssignments of G. In other wors, there exist lique trees T n T of G suh tht τ = ε T n τ = ε T. Let T T = ( ) T,{T u } u V(G) n TT = ( T,{T u} ) u V(G) be the orresponing tree moels of G. In other wors, for eh u V(G), we hve T u = T [ {C V(T) u C} ] n T u = T [ {C V(T ) u C} ]. In ition, for eh u V(G) n eh mximl lique C of G, efine the multisets τ u (C) = {S S τ(c), u S} n τ u(c) = {S S τ (C), u S}. Now, to prove the theorem, it suffies to emonstrte tht L(T u ) L(T u) for every u V(G). Consier u V(G) n efine two sequenes of integers 1,..., k n b 1,..., b k where i = τ u (C i ) n b i = τ u(c i ) for { ll i {1,..., k}. Note tht τ u (C) = τ u(c) for ll C {C 1,..., C k }, n by Lemm 10, we hve L(T u ) = C τ u (C) = 1 } n L(T u) = { C τ u(c) = 1 }. This implies the following. { } { L(T u ) L(T u) = C i τu (C i ) = 1 C i τ {i u (C i ) = 1} = i = 1} {i b i = 1} In other wors, the proof boils own to showing tht{i b i = 1} oes not hve more elements thn{i i = 1}. Rell tht, by the efinition of the ugmenting pth, τ(c k ) = 1 n τ(c i ) = 2 for ll i {2,..., k 1}. Notbly, sine the pth is shortest, C 1,..., C k re istint mximl liques of G. Thus, s τ u (C) τ(c) for ll C, we onlue tht k 1 n i 2 for ll i {2,..., k 1}. Further, note tht τ (C k ) = 2 while τ (C i ) = τ(c i ) = 2 for ll i {2,..., k 1}. In other wors, we hve b i 2 for ll i {2,..., k}. 11

12 We shll use the following two lims to show tht {i b i = 1} {i i = 1}. (6) If b i = 1, then i 1. Consier i {1,..., k} suh tht b i = 1, n ssume for ontrition tht i = 0. Sine b i = 1, we hve by Lemm 10 tht 1 = b i = τ u(c i ) = eg T u (C i ). In other wors, C i is lef of T u, n thus T u ontins t lest two verties. Rell tht V(T u ) = V(T u), n note tht 0 = i = τ u (C i ) = eg Tu (C i ) by Lemm 10. This mens tht C i is vertex of T u with no neighbour in T u. This is lerly impossible, sine T u is onnete n V(T u ) = V(T u) 2. Thus we must onlue tht i 1. This proves (6). (7) If b i = 1 n i 2, then there exists j > i suh tht j = 1, b j = 2, n r = b r for ll r {i+1,..., j 1}. To see this, first rell the onstrution of τ from τ by moving the tokens S 1,..., S k 1 s follows. ( τ(c i )\{S i } ) if i = 1 τ (C i ) = τ(c i )\{S i } {S i 1 } if 1 < i < k τ(c i ) {S i 1 } if i = k Also rell tht i = τu (C i ) = { S S τ(c i ), u S } n bi = τ u (C i ) = { S S τ (C i ), u S }. From these two fts we onlue the following reltionship between the vlues of i n b i (1 < i < k). { i if u S i S i 1 { 1 1 if u S ( ) b 1 = 1 b 1 if u S i = i 1 if u S i \ S i 1 k + 1 if u S b 1 i + 1 if u S i 1 \ S k = k 1 i k if u S k 1 i if u S i 1 S i Now, for the proof of (7), onsier i {1,..., k} suh tht b i = 1 n i 2. By ( ), we hve b i i 1 n thus i = 2. Further, i < k sine b k k by ( ), but b i = 1 < 2 = i. Moreover, u S i by ( ), sine i < k n b i = i 1. We let j be the lrgest in{i+ 1,..., k+1} suh tht r = b r for eh r {i+ 1,..., j 1}. First, we observe tht u S r for eh r {i,..., j 2}. Inee, if otherwise, we let r be the smllest inex in {i,..., j 2} with u S r. As we just rgue, we hve u S i, n so r > i. Therefore, u S r 1 by the minimlity of r. But then b r = r + 1 by ( ), sine 1 i < r < j 1 k, ontrition. This lso implies tht j k. Inee, if j = k+1, then i j 2 = k 1 sine i < k. Thus u S j 2 = S k 1 whih yiels b k = k + 1 by ( ). However, k {i+1,..., j 1} n so b k = k by the hoie of j. We n now onlue tht u S j 1. Inee, if i = j 1, then we use the ft tht u S i. Otherwise, i j 2 in whih se u S j 2 s rgue bove, n thus u S j 1 by ( ), sine j 1 = b j 1 n 1 i < j 1 < k. Finlly, we onsier the vlue of j. First, suppose tht j = k. Then b k = k + 1, sine u S j 1 = S k 1. We rell tht k 1 n so b k {1, 2}. If b k = 1, we hve k 1 by (6), but then k b k = k + 1 > k, ontrition. So, we must onlue b k = 2 n k = 1. Thus, s j = k, we hve b j = 2, j = 1, n r = b r for ll r {i+ 1,..., j 1} s require. Thus we my ssume tht j < k. By the mximlity of j, we hve j = b j. Also, u S j 1 n 1 i < j < k. So by ( ) we onlue tht b j = j + 1. We rell tht b j 2 s j > 1. Thus b j {1, 2} s j 0. Agin, if b j = 1, we onlue j 1 by (6) in whih se j b j > j, ontrition. Thus b j = 2, j = 1, n r = b r for ll r {i+ 1,..., j 1}, s require. This proves (7). We re now rey to onlue the proof. Denote A = {i i = 1} n B = {i b i = 1}. We show tht B A whih will imply the present theorem s rgue bove the lim (6). For eh i B, if i = 1, efine ϕ(i) = i. Otherwise, efine ϕ(i) = j where j is the inex obtine by pplying (7) for i. Note tht j = 1 n b j = 2. It follows tht ϕ is mpping from B to A. We show tht ϕ is, in ft, n injetive mpping. Suppose otherwise, n let i, i + be istint elements of B suh tht ϕ(i) = ϕ(i + ). Rell tht b i = b i + = 1 n note tht i ϕ(i) n i + ϕ(i + ). If i = ϕ(i), then i + ϕ(i + ) = ϕ(i) = i implying i + < ϕ(i + ) s i n i + re istint. So i + = 1 by the efinition of ϕ, n hene b ϕ(i + ) = 2 s ϕ(i + ) ws obtine by pplying (7) for i +. But then 1 = b i = b ϕ(i) = b ϕ(i + ) = 2, ontrition. Thus we must onlue tht i < ϕ(i) n, by symmetry, lso i + < ϕ(i + ). Now, without loss of generlity, ssume i < i +. Sine i + < ϕ(i + ), we must hve i + = 1 by the efinition of ϕ. However, b i + = 1 s i + B, n hene, i + = b i +. Rell tht the hoie of ϕ(i) using (7) for i gurntees tht r = b r for ll r {i+ 1,..., ϕ(i) 1}. 12

13 In prtiulr, i < i + < ϕ(i + ) = ϕ(i) n so i + = b i + whih is ontrition. This verifies tht ϕ is inee n injetive mpping from B to A, whih yiels B A. This ompletes the proof of Theorem Conluing Remrks In this pper we hve stuie the vertex lefge of horl grphs. Speifilly, horl grph G = (V, E) hs vertex lefge k when it hs tree moel ( T,{T u } u V ) suh tht eh subtree Tu hs t most k leves. We hve shown tht, for every fixe k 3, it is NP-omplete to eie if split grph G hs vertex lefge t most k, even when G is known to hve vertex lefge t most k+1. Moreover, we hve prove tht it is NP-hr to fin tree moel of G with s few leves in subtrees in totl s possible, n it is lso NP-hr to fin lique tree where s mny subtrees s possible re pths, even if G is split grph of vertex lefge 3. Interestingly, this mkes the polynomil-time reognition of pth grphs [2, 8, 18] the only trtble unprmeterize se of this problem. On the positive sie, we hve emonstrte n n O(l) lgorithm to ompute the vertex lefge of horl grph whose lefge is boune byl. This puts vertex lefge in the lss XP when the lefge is tken s the prmeter. Finlly, we hve shown tht every horl grph G hs tree moel whih simultneously relizes G s lefge n vertex lefge. In proving this result we hve lso shown tht, for every pth grph G, there exists pth moel withl(g) leves in the host tree n suh pth moel n be ompute in O(n 3 ) time. The following questions remin open. (A) for prmeter p (integer) is ny of the following problems in XP: given horl grph G is there tree moel of G where t most p subtrees re pths? given horl grph G is there tree moel of G where the totl number of leves in ll subtrees is t most p? (B) if the nswer to (A) is ffirmtive, is the problem in question in FPT or is it W[t]-hr for some (ll) t? (C) is the vertex lefge FPT with respet to lefge? (D) is the vertex lefge FPT with respet to some other grph prmeter? Aknowlegement We woul like to thnk nonymous referees for thoroughly reing the mnusript n for proviing helpful omments. We woul lso like to knowlege Mris Gutierrez n Pblo e Cri whose work [4] n lter isussions inspire the proofs of Theorems 3 n 4. The initil work on this projet ws one uring the seon uthor s visit t the University of Toronto in 2009 n lter uring the first uthor s visit t the Cesre Rothshil Institute of the University of Hif in Both trips were me possible by generous support of Prof. Derek Corneil of the University of Toronto vi his NSERC grnt. The seon uthor lso grtefully knowleges support from EPSRC, wr EP/I01795X/1. Referenes [1] BUNEMAN, P. A hrteriztion of rigi iruit grphs. Disrete Mthemtis 9 (1974), [2] CHAPLICK, S. Pth Grphs n PR-trees. PhD thesis, University of Toronto, [3] COHEN, E., GOLUMBIC, M., LIPSHTEYN, M., AND STERN, M. Wht is between horl n wekly horl grphs? In Grph-Theoreti Conepts in Computer Siene (WG 2008), Leture Notes in Computer Siene 5344 (2008), pp [4] DE CARIA, P., AND GUTIERREZ, M. Determining possible sets of leves for spnning trees of ully horl grphs. In MACI 2009: Congreso e Mtemáti Apli, Computionl e Inustril (2009), pp [5] FULKERSON, D. R., AND GROSS, O. A. Iniene mtries n intervl grphs. Pifi Journl of Mthemtis 15 (1965), [6] GAREY, M. R., AND JOHNSON, D. S. Computers n Intrtbility: A Guie to the Theory of NP-Completeness. W. H. Freemn & Co., New York, NY, USA, [7] GAVRIL, F. The intersetion grphs of subtrees of trees re extly the horl grphs. Journl of Combintoril Theory Series B 16 (1974), [8] GAVRIL, F. A reognition lgorithm for the intersetion grphs of pths in trees. Disrete Mthemtis 23 (1978), [9] GAVRIL, F. Generting the mximum spnning trees of weighte grph. Journl of Algorithms 8 (1987), [10] GOLUMBIC, M. C., LIPSHTEYN, M., AND STERN, M. Equivlenes n the omplete hierrhy of intersetion grphs of pths in tree. Disrete Applie Mthemtis 156 (2008),

14 [11] HABIB, M., AND STACHO, J. Liner lgorithms for horl grphs of boune irete vertex lefge. Eletroni Notes in Disrete Mthemtis 32 (2009), [12] HABIB, M., AND STACHO, J. Polynomil-time lgorithm for the lefge of horl grphs. In Algorithms - ESA 2009, Leture Notes in Computer Siene 5757 (2009), pp [13] JAMISON, R. E., AND MULDER, H. M. Constnt tolerne intersetion grphs of subtrees of tree. Disrete Mthemtis 290 (2005), [14] LÉVÊQUE, B., MAFFRAY, F., AND PREISSMANN, M. Chrterizing pth grphs by forbien inue subgrphs. Journl of Grph Theory 62 (2009), [15] LIN, I. J., MCKEE, T. A., AND WEST, D. B. The lefge of horl grph. Disuss. Mth. Grph Theory 18 (1998), [16] MONMA, C. L., AND WEI, V. K.-W. Intersetion grphs of pths in tree. Jounl of Combintoril Theory B 41 (1986), [17] SCHAEFER, T. J. The omplexity of stisfibility problems. In STOC (1978), pp [18] SCHÄFFER, A. A. A fster lgorithm to reognize unirete pth grphs. Disrete Applie Mthemtis 43 (1993), [19] SHIBATA, Y. On the tree representtion of horl grphs. Journl of Grph Theory 12 (1988), [20] TARJAN, R. E. Depth first serh n liner grph lgorithms. SIAM Journl on Computing 1 (1972), [21] WALTER, J. R. Representtions of horl grphs s subtrees of tree. Journl of Grph Theory 2 (1978),